Answer:
![\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%200.8708%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D)
Explanation:
The adiabatic throttling process is modelled after the First Law of Thermodynamics:
![m\cdot (h_{in} - h_{out}) = 0](https://tex.z-dn.net/?f=m%5Ccdot%20%28h_%7Bin%7D%20-%20h_%7Bout%7D%29%20%3D%200)
![h_{in} = h_{out}](https://tex.z-dn.net/?f=h_%7Bin%7D%20%3D%20h_%7Bout%7D)
Properties of water at inlet and outlet are obtained from steam tables:
State 1 - Inlet (Liquid-Vapor Mixture)
![P = 1500\,kPa](https://tex.z-dn.net/?f=P%20%3D%201500%5C%2CkPa)
![T = 198.29\,^{\textdegree}C](https://tex.z-dn.net/?f=T%20%3D%20198.29%5C%2C%5E%7B%5Ctextdegree%7DC)
![h = 2726.9\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h%20%3D%202726.9%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
![s = 6.3068\,\frac{kJ}{kg\cdot K}](https://tex.z-dn.net/?f=s%20%3D%206.3068%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D)
![x = 0.967](https://tex.z-dn.net/?f=x%20%3D%200.967)
State 2 - Outlet (Superheated Vapor)
![P = 200\,kPa](https://tex.z-dn.net/?f=P%20%3D%20200%5C%2CkPa)
![T = 130\,^{\textdegree}C](https://tex.z-dn.net/?f=T%20%3D%20130%5C%2C%5E%7B%5Ctextdegree%7DC)
![h = 2726.9\,\frac{kJ}{kg}](https://tex.z-dn.net/?f=h%20%3D%202726.9%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D)
![s = 7.1776\,\frac{kJ}{kg\cdot K}](https://tex.z-dn.net/?f=s%20%3D%207.1776%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D)
The change of entropy of the steam is derived of the Second Law of Thermodynamics:
![\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%207.1776%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D%20-%206.3068%5C%2C%20%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D)
![\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%200.8708%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20K%7D)
Answer:
The answer is below
Explanation:
Given that:
Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength
, Fracture strength
![(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa](https://tex.z-dn.net/?f=%28%5Csigma_f%29%3D5100%5C%20MPa%3D5100%2A10%5E6%5C%20Pa%2Cfiber-matrix%5C%20stres%28%5Csigma_m%29%3D17.5%5C%20MPa%3D17.5%2A10%5E6%5C%20Pa%2Cmatrix%5C%20strength%3D%5Ctau_c%3D17%5C%20MPa%3D17%20%2A10%5E6%5C%20Pa)
a) The critical length (
) is given by:
![L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm](https://tex.z-dn.net/?f=L_c%3D%5Csigma_f%2A%28%5Cfrac%7BD%7D%7B2%2A%5Ctau_c%7D%20%29%3D5100%2A10%5E6%2A%5Cfrac%7B0.00003%7D%7B2%2A17%2A10%5E6%7D%3D0.0045%5C%20m%3D4.5%5C%20mm)
The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.
b) The volume fraction (Vf) is gotten from the formula:
![\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}} \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456](https://tex.z-dn.net/?f=%5Csigma_%7Bcd%7D%3D%5Cfrac%7BL%2A%5Ctau_c%7D%7BD%7D%2AV_f%2B%5Csigma_m%281-V_f%29%5C%5C%5C%5CV_f%3D%5Cfrac%7B%5Csigma_%7Bcd%7D-%5Csigma_%7Bm%7D%7D%7B%5Cfrac%7BL%2A%5Ctau_c%7D%7BD%7D-%5Csigma_%7Bm%7D%7D%20%20%5C%5C%5C%5CSubstituting%3A%5C%5C%5C%5CV_f%3D%5Cfrac%7B630%2A10%5E6-17.5%2A10%5E6%7D%7B%5Cfrac%7B0.0024%2A17%2A10%5E6%7D%7B0.00003%7D%20-17.5%2A10%5E6%7D%20%5C%5C%5C%5CV_f%3D0.456)
On highways, the far left lane is usually the<u> fastest</u> moving traffic.
Answer: Option D.
<u>Explanation:</u>
For the most part, the right lane of a freeway is for entering and leaving the traffic stream. It is an arranging path, for use toward the start and end of your interstate run. The center paths are for through traffic, and the left path is for passing. On the off chance that you are not passing somebody, try not to be driving in the left path.
Regular practice and most law on United States expressways is that the left path is saved for passing and quicker moving traffic, and that traffic utilizing the left path must respect traffic wishing to surpass.
Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
=
...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
=
= 12 kpsi
and
=
...........2
put here value and we get
=
= 17.34 kpsi
now we apply here goodman line equation here that is
...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
solve it we get
FOS = 1.5432